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PHY 201
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A ball is tossed vertically upward and caught at the position from which it was released.
• Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?
answer/question/discussion: ->->->->->->->->->->->-> :
if I caught it at its exact initial position I would imagine that it would be faster since its initial velocity was zero. But that thinking is obviously wrong. If I apply a force to move the ball greater than the force of gravity on the way back down, then it would be slower then the speed it was initially launched. It just feels like when I toss a ball up in the air it rises faster than it falls, but that may just be an illusion because I have not done any experiments.
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Gravity will exert exactly the same forces at the same positions on the way up as on the way down.
By assumption, no other forces will act on the ball (between release and catch your hand is not in contact with the ball and cannot exert a force on it).
So how will the work done by gravity on the way up compare to the work done be gravity on the way down?
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• What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion: ->->->->->->->->->->->-> :
If air resistance is present, I guess that the resistance would work against the ball more on its way down then up, or about the same. The ball would leave and return slower then the ball tossed up with no air resistance.
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Will air resistance do positive or negative work on the ball as it rises? What about as it falls?
From release to catch, then, does air resistance do positive or negative work on the ball?
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Good start, but let's try to clarify your thinking a little.
Can you revise by answering the questions I've posed?
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